Integrate: x times the square root of 2 - 3x Dx.?

Question

Im not able to put the Integral sign or the square root sign. However, the problem reads exactly as it is.

Hint: a simple u substitution will not work.

I will award best answer to the first person who has the CORRECT answer!

Using integration by parts will not work either. Look in a table of integrals.

Nobody has gotten the CORRECT answer yet. Here is the answer, show me the Identity you used and your work.

answer = -2/135(9x + 4)(2 - 3x)^3/2 + C

Answer 1

x*sqrt(2-3x)dx
A substitution will work! (where did you get the hint that it will not work?--that may have mislead you)
Let u = 2-3x, then du = -3 dx or dx = (-1/3) du
Now since u = 2 - 3x, you can solve for x.
u = 2 - 3x
u - 2 = -3x
(u-2)/-3 = x or
x = (-1/3)(u-2)
Now the integral is
(-1/3)(u-2)sqrt(u)(-1/3) du
= (1/9)(u-2)(u^(1/2))du
= (1/9)(u^(3/2) - 2u^(1/2)) du
= (1/9)[(2/5)u^(5/2) - (4/3)u^(3/2)] + C
= (1/9)[(2/5)(2-3x)^(5/2) - (4/3)(2-3x)^(3/2)] + C
You probably don't need to distribute the 1/9, but you can if you think it is necessary.
I would not use trig substitution. It is not necessary.

Answer 2

I got it, let me type it up in Tex so that it is easily read, and then I'll send you a link to the file.


Sorry it took so long to type up:

http://www.geocities.com/euleratuo/
(add C to the end, always forget the constant :))


and you of course should be able to differentiate to prove it is correct

hehe, yeah regular substitition does work . . . maybe I should have tried that first :) Been a while since I have actually integrated by hand (needed the practice). Thanks Mathgirl

Answer 3

int(x*sqrt(2-3x)*dx) right?

This is definitely a trig substitution. Rewrite:

int(x^((1/2)*2)*sqrt(1-(sqrt(3)/sqrt(2)^2)^2*sqrt(x)^2)*dx)

int(x^((1/2)*2)*sqrt(1-(sqrt(3x/2)^2)*dx)

This probably doesn't make much sense, but now you can use trig identities to simplify.

Use sin(theta)=sqrt(3x/2)

Go from there.

Answer I got:

(1/10)*(3-2x)^(5/2) - (1/2)*(3-2x)^(3/2) + C

And I rechecked it using derivatives. :P

Answer 4

int(x*sqrt(2-3x)dx)
let:
u=x
du=dx
dv=sqrt(2-3x)dx
v=-(2/9)(2-3x)^(3/2)

therefore:
int(x*sqrt(2-3x)dx) = uv - int(vdu)
= -(2x/9)(2-3x)^(3/2) + int[(2/9)(2-3x)^(3/2)dx]
= -(2x/9)(2-3x)^(3/2) - (4/135)(2-3x)^(5/2) + C ; where c is the arbitrary constant

Answer 5

using u-v substitution:

u=sqrt(2-3x)
dv=x*dx
v=.5*x^2
du=-1.5/sqrt(2-3x)

integral(u*dv) = u*v - integral(v*du)

ANS: sqrt(2-3x)*.5*x^2 + integral((.75*x^2)/sqrt(2-3x)) + C

(C = unknown constants)